OFFSET
1,2
COMMENTS
Numbers for which the sum of distinct prime factors equals the sum of exponents in the prime factorization, A008472(n)=A001222(n). - Reinhard Zumkeller, Mar 08 2002
LINKS
Giuseppe Coppoletta and Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 100 terms from G. Coppoletta)
EXAMPLE
320 is included because 320 = 2^6 * 5^1 and 2+5 = 6+1.
MATHEMATICA
f[n_]:=Plus@@First/@FactorInteger[n]==Plus@@Last/@FactorInteger[n]; lst={}; Do[If[f[n], AppendTo[lst, n]], {n, 0, 3*8!}]; lst (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
max = 10^12; Sort@Reap[Sow@1; Do[p = Select[IntegerPartitions[se, All, Prime@ Range@ PrimePi@ se], Sort[#] == Union[#] &]; Do[ np = Length[f]; va = IntegerPartitions[se, {np}, Range[se]]; Do[pe = Permutations[v]; Do[z = Times @@ (f^e); If[z <= max, Sow@z], {e, pe}], {v, va}], {f, p}], {se, 2, Log2[max]}]][[2, 1]] (* Giovanni Resta, May 07 2016 *)
PROG
(PARI) for(n=1, 10^6, if(bigomega(n)==sumdiv(n, d, isprime(d)*d), print1(n, ", ")))
(PARI) is(n)=my(f=factor(n)); sum(i=1, #f~, f[i, 1]-f[i, 2])==0 \\ Charles R Greathouse IV, Sep 08 2016
(Sage) def d(n):
v=factor(n)[:]; L=len(v); s0=sum(v[j][0] for j in range(L)); s1=sum(v[j][1] for j in range(L))
return s0-s1
[k for k in (1..100000) if d(k)==0] # Giuseppe Coppoletta, May 07 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, May 09 2000
STATUS
approved